Solved Problems in Classical Electromagnetism

Solved Problems in Classical Electromagnetism

by Jerrold Franklin

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Overview

Solved Problems in Classical Electromagnetism is a valuable tool to help students learn to do physics while using concepts they learn in the courses. Students who are taking or have already taken an advanced EM course will find the book to be a useful adjunct to their textbook, giving added practice in applying what they are learning. For students who are taking an undergraduate EM course and want to get more depth, this book can help them achieve that aim and also help them prepare for graduate work. Beginning students, or those not even taking a course at the moment, can benefit from these problems and learn just from working on them with the help of the solutions. In each chapter, the problems start out relatively easy and then get progressively more advanced, helping students to go just as far as they can at their present level.
The book includes a number of review sections to assist students without previous advanced training in working out the problems. The first review section is a comprehensive development of vector calculus that will prepare students to solve the problems and provide a strong foundation for their future development as physicists. The problems are drawn from the topics of electrostatics, magnetostatics, Maxwell's equations, electromagnetic radiation, and relativisitc electromagnetism.

Product Details

ISBN-13: 9780486813721
Publisher: Dover Publications
Publication date: 09/12/2018
Series: Dover Books on Physics
Pages: 256
Product dimensions: 6.14(w) x 9.21(h) x (d)

About the Author


Jerrold Franklin is Emeritus Professor of Physics at Temple University.

Read an Excerpt

CHAPTER 1

Electrostatics

1.1 Static fields and forces

1.1.1 Applications of Coulomb's law

1. Four point charges, each of charge q and mass m, are located at the four corners of a square of side L.

(a) Find the magnitude of the force on one of the charges.

(b) Use the force in part (a) to find the velocity of one of the charges a long time after the four charges are released from rest in the original configuration.

Solution:

(a) The force on the upper right-hand charge due to the other three charges at the corners of a square with sides of length L is

[MATHEMATICAL EXPRESSION OMITTED] (1.1)

The magnitude of this force on any of the four charges is

F = q2(1 + 2 [square root of 2]/2L2. (1.2)

(b) If the four charges are released from rest, we can write for the acceleration of any of the charges

F/m = a = dv/dt = dv/dL' dL'/dt, (1.3)

where L' is the side length of the square at any time during the motion. The center of the square remains fixed, and the distance, r, of a charge from the center is related to L' by L' = [square root of 2]r, so

[MATHEMATICAL EXPRESSION OMITTED] (1.4)

and the squared velocity after a long time is given by

[MATHEMATICAL EXPRESSION OMITTED] (1.5)

The velocity is the square root of this:

V = q [(1 + 2 [square root 2])/[square root of 2]mL]1/2. (1.6)

2. Four point charges, each of charge q and mass m, are located at the four corners of a square of side L.

(a) Find the potential energy of this configuration.

(b) Use conservation of energy to find the velocity of one of the charges a long time after the four charges are released from rest in the original configuration.

Solution:

(a) In the configuration of four point charges q at the four corners of a square of side L, each of the four charges is a distance L from two other charges, and a distance v2L from a third charge. Consequently, the potential energy is given by

[MATHEMATICAL EXPRESSION OMITTED] (1.7)

(b) The kinetic energy of the four charges (each of mass m) at a long time after their release (so that L -> ∞) is given by

[MATHEMATICAL EXPRESSION OMITTED] (1.8)

[MATHEMATICAL EXPRESSION OMITTED] (1.9)

3. Four point charges, each of charge q, are fixed at the four corners of a square of side L. Find the electric field a distance z above the plane of the square on the perpendicular axis of the square.

Solution:

The four point charges q are located at the corners of a square with sides of length L. The distance from each charge to a point z above the square, on the perpendicular axis of the square, is [square root of z2 + L2/2]. The horizontal fields cancel, and the magnitude of the vertical field is given by

[MATHEMATICAL EXPRESSION OMITTED] (1.10)

4. Show that the electric field a distance r from a long straight wire with a uniform linear charge density λ is given by

E = 2λ[??]/r. (1.11)

Solution:

By symmetry, the electric field of a long straight wire is perpendicular to the wire. The field a distance r from the wire is given by

[MATHEMATICAL EXPRESSION OMITTED] (1.12)

We made the substitution z = r tan θ in doing the integral.

5. Two long parallel wires, each with uniform linear charge density λ, are a distance a apart. The origin of coordinates is the midpoint between the two wires.

(a) Find the electric field in terms of the vectors r and a.

(b) Write down the x and y components of E, taking a in the x direction.

Solution:

(a) For the con?guration of two wires a distance a apart, the electric field is

[MATHEMATICAL EXPRESSION OMITTED] (1.13)

(b) The electric field in Cartesian coordinates is

[MATHEMATICAL EXPRESSION OMITTED] (1.14)

[MATHEMATICAL EXPRESSION OMITTED] (1.15)

6. A straight wire of length L has a uniformly distributed charge Q. Find the electric field on the axis of the wire a distance z from the center of the wire for the cases

(a) z > L/2.

(b) -L/2 < z< L/2. (Hint: Use symmetry for part (b).)

Solution:

(a) The field on the axis of the uniformly charged wire, a distance z from the center of the wire, is given for z >L/2 by

[MATHEMATICAL EXPRESSION OMITTED] (1.16)

(b) For -L/2 < z< L/2, the point z is a distance (L/2 - z) from the end of the wire. The wire can be thought of as two parts.

The part of the wire from z' = z - (L/2 - z) = 2z - L/2 to z' = L/2 is symmetric about the point z. This means that the field due to that portion of the wire will cancel. The remaining part of the wire has a length L' = L - 2(L/2 - z) = 2z and a charge Q' = 2zQ/L. The midpoint of this part of the wire is at z0 = (2z - L/2 - L/2)/2 = z - L/2. Thus the electric field from this part of the wire is

[MATHEMATICAL EXPRESSION OMITTED] (1.17)

7. (a) Find the electric field a distance z along the axis of a uniformly charged ring of charge Q and radius R.

(b) Integrate the field for a ring to find the electric field a distance z on the axis of a uniformly charged disk of charge Q and radius R.

(c) Find the electric field of the charged disk for

(1) z = 0+ (just above the disk).

(2) z » R. (Expand the square root.)

Solution:

(a) Every point on the uniformly charged ring is the same distance from a point a distance z along the axis of the ring, and a line from any point on the ring makes the same angle θ with the z-axis. Thus the electric field at z is

[MATHEMATICAL EXPRESSION OMITTED] (1.18)

(b) The disk has a surface charge density σ = QR2. It can be considered as a collection of rings, each of radius r' with a charge

dq = 2πr'σdr' = 2Qr'dr'/R2. (1.19)

The electric field a distance z along the axis of the disk is given as (using part (a))

[MATHEMATICAL EXPRESSION OMITTED] (1.20)

(c) (i) For z = 0+,

Ez = 2Q/R2 = 2πσ. (1.21)

(ii) For z » R, we write E as

[MATHEMATICAL EXPRESSION OMITTED] (1.22)

Using the binomial theorem, we get

[MATHEMATICAL EXPRESSION OMITTED] (1.23)

This limit, equal to the field of a point charge Q, can be used as a check on the original result.

8. A straight wire of length L has a uniformly distributed charge Q. Find Ex and Ey a distance d from the wire in the configuration shown in the figure on the next page. Express your answers in terms of the angles θ1 and θ2.

Solution:

The parallel component of the electric field a distance d from a uniformly charged straight wire of length L is given by the integral

[MATHEMATICAL EXPRESSION OMITTED] (1.24)

where x1 and x2 are the two endpoints of the wire. Let

[MATHEMATICAL EXPRESSION OMITTED] (1.25)

Then

[MATHEMATICAL EXPRESSION OMITTED] (1.26)

where θ1 and θ2 are the angles shown.

For the perpendicular component of E, the same substitution for z, leads to

[MATHEMATICAL EXPRESSION OMITTED] (1.27)

9. Four point charges, each of charge q, are fixed at the four corners of a square of side L.

(a) Find the potential a distance z above the plane of the square on the perpendicular axis of the square.

(b) Use the potential to find the electric field at the same point.

Solution:

(a) Each charge q, at the corner of a square with sides of length L, is a distance [square root of z2 + L2/2] from a point z along the perpendicular axis of the square. (See the figure in Prob. 3.)

Thus the potential at the point z is

φ = 4q/[square root of z2 + L2/2]. (1.28)

(b) The electric field is

[MATHEMATICAL EXPRESSION OMITTED] (1.29)

10. A straight wire of length L has a uniformly distributed charge Q.

(a) Find the potential on the axis of the wire a distance z > L/2 from the center of the wire. (b) Use the potential to find the electric field at the same point.

Solution:

(a) The potential on the axis of the uniformly charged wire, a distance z from the center of the wire, is given for z > L/2 by

[MATHEMATICAL EXPRESSION OMITTED] (1.30)

(b) The electric field is

[MATHEMATICAL EXPRESSION OMITTED] (1.31)

11. (a) Find the potential a distance z along the axis of a uniformly charged ring of charge Q and radius R.

(b) Integrate the potential for a ring to find the potential a distance z along the axis of a uniformly charged disk of charge Q and radius R.

(c) Use the potentials in parts (a) and (b) to find the corresponding electric fields.

Solution:

(a) Every point on the uniformly charged ring is the same distance [square root of z2 + R2] from a point a distance z along the axis of the ring, so the potential at z is

φ = Q/[square root of z2 + R2]. (1.32)

(b) The uniformly charged disk of radius R has a surface charge density σ = QR2. It can be considered as a collection of rings, each of radius r' with a charge

dq = 2πr'σdr' = 2Qr'dr'/R2. (1.33)

The potential a distance z along the axis of the disk is given as (using part (a))

[MATHEMATICAL EXPRESSION OMITTED] (1.34)

(c) The electric field of the ring is

[MATHEMATICAL EXPRESSION OMITTED] (1.35)

The electric field of the disk is

[MATHEMATICAL EXPRESSION OMITTED] (1.36)

12. (a) Integrate the potential for a ring to find the potential a distance d from the center of a uniformly charged spherical shell of charge Q and radius R.

(b) Integrate the potential for a spherical shell to find the potential a distance d > R from the center of a uniformly charged sphere of charge Q and radius R.

(c) Use the potentials in parts (a) and (b) to find the corresponding electric fields.

Solution:

(a) A uniformly charged spherical shell has a surface charge density

σ = Q/4πR2. (1.37)

The spherical surface can be thought of as composed of rings of radius r = R sin θ, where θ is the angle from the z-axis. Each ring has a charge

dq = 2πrσR dθ = 1/2 Q sin θ dθ. (1.38)

The distance from the plane of a ring to the point d is z = d - R cosθ. Thus the potential at a point dR from the center of the sphere is

[MATHEMATICAL EXPRESSION OMITTED] (1.39)

The potential for dR is

[MATHEMATICAL EXPRESSION OMITTED] (1.40)

For d outside the sphere, the potential is the same as that of a point charge. For d inside the sphere, the potential is constant and hence the electric field is zero.

(b) The potential at a distance dR from the center of a uniformly charged spherical shell is Δq/d, where Δq is the charge on the shell. The potential does not depend on the radius R of the shell, as long as dR. A uniformly charged sphere can be considered a collection of the uniformly charged shells, so that the potential due to the uniformly charged sphere will be φ = Q/d, where Q is the net charge on the sphere. (Q is the sum of all the Δq on each spherical shell.)

Note that the charged sphere need not be uniformly charged as long as its charge distribution is spherically symmetric.

(c) Since the potentials in a and b are the same as the potential of a point charge, the electric fields are the same as the electric field of a point charge:

E = Q[??]/r2, r ≥ R. (1.41)

for rR.

1.1.2 Gauss's law

13. (a) Use Gauss's law to find the electric field inside and outside a uniformly charged hollow sphere of charge Q and radius R.

(b) Integrate E to find the potential inside and outside the hollow sphere.

Solution:

(a) By symmetry, the electric field of the uniformly charged hollow sphere is in the radial direction. We consider a Gaussian sphere of radius r, concentric with the hollow sphere. Then, for rR, Gauss's law becomes

[MATHEMATICAL EXPRESSION OMITTED] (1.42)

For rR, Gauss's law is

[MATHEMATICAL EXPRESSION OMITTED] (1.43)

(b) The electric field for rR is the same as that for a point charge, so the potential is the same as the potential of a point charge:

φ(r) = Q/r. (1.44)

For rR, E = 0, so the interior of a uniformly charged shell is an equipotential with

φ = Q/R. (1.45)

14. Find the potential energy of a uniformly charged shell of charge Q and radius R by integrating

(a) ρφ.

(b) E2.

Solution:

(a) The potential energy of a uniformly charged spherical shell of charge Q and radius R is given by

[MATHEMATICAL EXPRESSION OMITTED] (1.46)

(b) The potential energy is also given by integrating E2:

[MATHEMATICAL EXPRESSION OMITTED] (1.47)

15. (a) Use Gauss's law to find the electric field inside and outside a uniformly charged solid sphere of charge Q and radius R.

(b) Integrate E to find the potential inside and outside the solid sphere.

Solution:

(a) The charge density inside the uniformly charged sphere is

ρ = 3Q/4πR3. (1.48)

By symmetry, the electric field is in the radial direction. We consider a Gaussian sphere of radius r, concentric with the uniformly charged sphere. Then, for rR, Gauss's law becomes

[MATHEMATICAL EXPRESSION OMITTED] (1.49)

For rR, Gauss's law is

[MATHEMATICAL EXPRESSION OMITTED] (1.50)

(b) The electric field for rR is the same as that for a point charge, so the potential is the same as the potential of a point charge:

φ(r) = Q/r, r ≥ R. (1.51)

For rR,

[MATHEMATICAL EXPRESSION OMITTED] (1.52)

16. Find the potential energy of a uniformly charged sphere of charge Q and radius R by integrating

(a) ρφ.

(b) E2.

Solution:

(a) The potential energy of a uniformly charged sphere of charge Q and radius R is given by (using the answer to problem 15(b))

[MATHEMATICAL EXPRESSION OMITTED] (1.53)

(b) The potential energy is also given by integrating E2:

[MATHEMATICAL EXPRESSION OMITTED] (1.54)

the same as in part (a).

(Continues…)


Excerpted from "Solved Problems in Classical Electromagnetism"
by .
Copyright © 2018 Jerrold Franklin.
Excerpted by permission of Dover Publications, Inc..
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Table of Contents


Preface. Electrostatics
2. Magnetostatics
3. Electromagnetism
4. EM Radiation
5. Relativistic Electromagnetism

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